Question

The radius of the least circle passing through the point (8,4) and cutting the circle $x^2+y^2=40$ orthologonally is

• A. $\sqrt{5}$
• B. $\sqrt{7}$
• C. $2\sqrt{5}$
• D. $4\sqrt{5}$

Answer 1

The correct option is A $\sqrt{5}$

Let the circle be $x^2+y^2+2gx+2fy+c=0$ .........$\left(1\right)$

Given circle is $x^2+y^2=40$ .......$\left(2\right)$

These two circles are orthogonal

Hence $2g_1{g}_2+2f_1{f}_2=c_1+c_2$

$⟹0+0=c-40$

$\therefore c=40$

$\left(1\right)$ passes through $(8,4)$

$64+16+16g+8f+40=0$

$\Rightarrow 120+16g+8f=0$

$\Rightarrow f+2g+15=0$

$\Rightarrow f=-(2g+15)$

radius$=\sqrt{g^2+f^2-c}=\sqrt{g^2+{(2g+15)}^2-40}$

For least circle radius must be minimum

Let $f\left(g\right)=g^2+{(2g+15)}^2-40$ is minimum.

$f^{\prime }\left(g\right)=2g+4(2g+15)=0$

$\Rightarrow 10g=-60$

$\Rightarrow g=-6$

$f^{\prime \prime }\left(g\right)=10>0$ minimum

$f=-(-12+15)=-3$

To minimise the radius, $g=-6$

Equation of the circle is $x^2+y^2-12x-6y+40=0$

Radius$=\sqrt{36+9-40}=\sqrt{5}$